Finding the work done using force vectors in 3 dimensions

AI Thread Summary
The discussion focuses on calculating the work done by a force of 3 Newtons acting in the direction of the vector 2i + j + 2k while moving an object 2 meters from (0, 0, 0) to (0, 2, 0). Participants clarify that the displacement vector is 2j and debate how to correctly represent the force vector. The initial confusion arises from the normalization of the direction vector and its relationship to the force's magnitude. Ultimately, it is established that the correct approach involves using the dot product of the force vector and the displacement vector, leading to the conclusion that the work done is 6 Joules. The discussion emphasizes the importance of accurately incorporating the force's magnitude and direction in vector calculations.
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Homework Statement


Find the work done by a force of 3 Newtons acting in the direction 2i + j + 2k in moving an object 2 meters from (0, 0, 0) to (0, 2, 0).


Homework Equations


Commonly known vector operations.


The Attempt at a Solution



I found the displacement vector to be 2j.
My problem is figuring out how to represent the direction and magnitude of force as an actual vector.
The answer is (6*SQRT 5)/5 Joules. I don't understand how.
 
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Are you sure that is the right answer?
 
Amok said:
Are you sure that is the right answer?

Positive. This really doesn't make any sense to me.
 
marthkiki said:

Homework Statement


Find the work done by a force of 3 Newtons acting in the direction 2i + j + 2k in moving an object 2 meters from (0, 0, 0) to (0, 2, 0).


Homework Equations


Commonly known vector operations.


The Attempt at a Solution



I found the displacement vector to be 2j.
My problem is figuring out how to represent the direction and magnitude of force as an actual vector.
The answer is (6*SQRT 5)/5 Joules. I don't understand how.

I think that "answer" was gotten using sqrt(2+1+2) to normalize the vector instead of correctly using sqrt(22+12+22). Typo in your answer book.
 
s
LCKurtz said:
I think that "answer" was gotten using sqrt(2+1+2) to normalize the vector instead of correctly using sqrt(22+12+22). Typo in your answer book.

That may well be true. If so, is the correct answer 2?
The displacement vector is 2j.

(2i + j + 2k) dot product 2j = 0 + 2j + 0 = 2j. So the correct answer is 2 Joules, correct?
 
That's what I got.
 
Wait where does the magnitude of 3 N come into play? Shouldn't it be multiplied to 2j to get the total work? So you get

W = F*AB
AB = 2j
F = 3(2i + j + 2k) = 6i + 3j + 6k

F dot product AB = 0 + 6j + 0 = 6j. So the work should in fact be 6 Joules, correct?
 
No, you only have to take into account the magnitude of the force in the (0,1,0) direction, because that is that component that is going to be doing all the work! Remember that:

WORK = FORCE dot product DISPLACEMENT
 
marthkiki said:
Wait where does the magnitude of 3 N come into play? Shouldn't it be multiplied to 2j to get the total work? So you get

W = F*AB
AB = 2j
F = 3(2i + j + 2k) = 6i + 3j + 6k

F dot product AB = 0 + 6j + 0 = 6j. So the work should in fact be 6 Joules, correct?

No. A force of 3N is represented by a vector of length 3. Your vector specifying the direction of the force just happens to have length 3 already. That is just a lucky break for this problem. Normally what you would do is multiply a unit vector in the direction of the force by 3.
 
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